Sec(x/2) = 1/cos(x/2)
sec(x/2)=cos(x/2) ----> cos^2(x/2)=1 ---> cos(x/2) = -1 and cos(x/2) = 1
Cos(x/2)=1 --- > x/2 = 0, only. x = 0;
cos(x/2)=-1 ----> x/2 = pi -> x = 2pi. But the statement says [0,2pi), so 2pi can not be chosen.
Only x = 0.
In fact, your equation is equivalent to sec(x)=cos(x), for x in [ 0, pi), so yes, only x = 0 .
Answer:
L=6x^2-4x+5
P=12x^2-4x-2
Step-by-step explanation:
A=WL
So, we know the area is 6x^2-8x-30 and the width is 2x-6
Substitute
(6x^2-8x-30)=(2x-6)(L)
L=(6x^2-8x-30)/(2x-6
Divide
L=6x^2-4x+5
P=2L+2W
Substitute
P=2(6x^2-4x+5)+2(2x-6)
P=12×^2-8x+10+4x-12
Simplify
P=12x^2-4x-2
Answer:
34
Step-by-step explanation:
(10)(4)−(4)(3)−2x−6x+8+(−3)(4)+(6)(3)−(2)(2)+8x−4
= 40+−12+−2x+−6x+8+−12+18+−4+8x+−4
Combine like terms:
= 40+−12+−2x+−6x+8+−12+18+−4+8x+−4
= (−2x+−6x+8x)+(40+−12+8+−12+18+−4+−4)
= 34
5(-1)-7(5)+(5)2
-5-35+10
-40+10
= -30
Answer:
length(l) is 8 m
area=56^2
l*b=56^2
8*b=56^2
b=56/8
hence,
breadth {width}(b) is 7 m
hence , perimeter is
2(l+b)
2(8+7)
2*15
30 cm
hence perimeter is 30 cm
